5
$\begingroup$

Task: find all values of the parameter, such that integral converges. $$\int_0^{+\infty} \frac{dx}{1+x^a \sin^2x}$$ I tried a lot and I used Cauchy and Weierstrass method but it was useless. And now I know that I must use $$\sum_n \int_{\pi n}^{\pi(n+1)}$$

$\endgroup$
3
$\begingroup$

First, we have $$ \small\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+((k+1)\pi)^\alpha\sin^2(x)} \le\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)} \le\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+(k\pi)^\alpha\sin^2(x)}\tag{1} $$ Then to estimate the bounds in $(1)$, we have $$ \begin{align} \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} &=\int_0^\pi\frac{\mathrm{d}x}{1+\beta\sin^2(x)}\\ &=2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)}\tag{2} \end{align} $$ Noting that on $[0,\pi/2]$, $\frac2\pi x\le\sin(x)\le x$, we can bound $(2)$ $$ 2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta x^2} \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\frac{4\beta}{\pi^2}x^2}\tag{3} $$ Evaluating the bounds in $(3)$, we get $$ \frac2{\sqrt\beta}\arctan\left(\frac{\pi\sqrt\beta}2\right) \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} \le\frac\pi{\sqrt\beta}\arctan\left(\sqrt\beta\right)\tag{4} $$ Combining $(1)$, $(2)$, and $(4)$, we get $$ \frac{\pi/2}{((k+1)\pi)^{\alpha/2}} \le2\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)} \le\frac{\pi^2/2}{(k\pi)^{\alpha/2}}\tag{5} $$ Therefore, as $k\to\infty$, $$ \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)}=\Theta\left(k^{-\alpha/2}\right)\tag{6} $$ Thus, the integral converges for $\alpha\gt2$.

$\endgroup$
1
$\begingroup$

In the following, we shall use the following result twice: $\int \limits _0 ^\pi {1 \over {1 + c \sin ^2 x}} = {\pi \over \sqrt {1 + c}}$, if $c>0$. To prove it, split the integral into $\int \limits _0 ^{\pi \over 2} + \int \limits _{\pi \over 2} ^\pi$ and use the substitution $t = \tan x$ on each interval.

Note that $\int \limits _0 ^\infty {1 \over 1+x^a \sin ^2 x} \mathbb d x \geq \int \limits _0 ^\infty {1 \over 1+x^a} \mathbb d x$, and the latter integral is divergent for $a \leq 1$, therefore necessarily $a>1$.

Next, note that $$\int \limits _{n \pi} ^{(n+1) \pi} {1 \over 1+x^a \sin ^2 x} \mathbb d x = \int \limits _0 ^\pi {1 \over 1 + (x+n \pi)^a \sin ^2 x} \mathbb d x \leq \int \limits _0 ^\pi {1 \over 1+(n \pi)^a \sin ^2 x} \mathbb d x= {\pi \over \sqrt {1 + (n \pi)^a}}$$ and, since $\sum {\pi \over \sqrt {1 + (n \pi)^a}}$ converges if and only if $a>2$, your given integral also is guaranteed to converge for $a>2$.

It remains to investigate the case $1 < a \leq 2$. But $$\int \limits _0 ^\pi {1 \over 1 + (x+n \pi)^a \sin ^2 x} \mathbb d x \geq \int \limits _0 ^\pi {1 \over 1 + (\pi+n \pi)^a \sin ^2 x} \mathbb d x = {\pi \over \sqrt {1 + \pi ^a (n+1) ^a}}$$ and, since the series $\sum {\pi \over \sqrt {1 + \pi ^a (n+1) ^a}}$ is divergent for $a \leq 2$, so will be your integral.

Therefore, the series converges if and only if $a>2$.

$\endgroup$
  • $\begingroup$ Just curious, why using the $\mathbb{d}$ symbol for the $dx$? $\endgroup$ – Dmoreno May 29 '15 at 17:59
  • $\begingroup$ @Dmoreno: Agreed, I myself have given it some thought before making this choice: I want to stress that the letter "d" stands for something else than a variable or a parameter. For the same reason, if you want, for which we prefix "sin", "log", "exp" etc. by a "\". I use the same formatting for the numbers "e" and "i", to distinguish them from variables. $\endgroup$ – Alex M. May 29 '15 at 18:07
  • $\begingroup$ Oh, I see. That's the same reason for which I always try to type $\mathrm{d}x$ instead of $dx$. Cheers! $\endgroup$ – Dmoreno May 29 '15 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.